(PS1 and PS2) Power Series#

In this lesson we are going to see:

  • how to find the radius of convergence for a power series.

  • how to find the interval of convergence for a power series.

Review Videos#

Power Series#

Power Series

A power series is a series of the form:

\[ \sum_{n=0}^{\infty} c_n x^n = c_0 +c_1 x+c_2x^2+ c_3 x^3+\cdots \]

More generally, a power series centered at a is a series of the form:

\[ \sum_{n=0}^{\infty} c_n (x-a)^n = c_0 +c_1 (x-a)+c_2(x-a)^2+ c_3 (x-a)^3+\cdots \]

Note that in this definition:

  • \(x\) is a variable

  • \(c_n\) is called a coefficient

  • we use the convention that \((x-a)^0=1\)

Domain of a Power Series#

Given a power series, we can define a function:

\[ f(x)=\sum_{n=0}^{\infty} c_n (x-a)^n \]

A natural question then is what numbers can we actually plug in for \(x\)? What is the domain of this function?

  • A number \(x\) is in the domain if the resulting series converges.

  • A number \(x\) is not in in the domain, if the resulting series diverges.

Example 1#

Find the domain of the following function:

\[ f(x)=\sum_{n=0}^{\infty} x^n \]
Solution (Click to see the steps.)

For each individual \(x\)-term, this power series is actually a geometric series where \(a=1\) and \(r=x\).

\[ \sum_{n=0}^{\infty} a r^n \]

We’ve previously looked at geometric series and know that they:

  • converge when \(|r|<1\)

  • diverge when \(|r|\geq 1\)

This means our power series converges when \(|x|<1\) or writing this as the open interval \((-1,1)\).

\[ |x|<1 \iff -1<x<1 \]

Therefore, the domain of this function is the interval \((-1,1)\).

Interval of Convergence#

Interval of Convergence

The Interval of Convergence for a power series is the interval of \(x\)-values for which the series is convergent. Essentially, this is the domain of the power series.

Given a power series \(\displaystyle \sum_{n=0}^{\infty} c_n (x-a)^n\) the Interval of Convergence can be one of the following:

\[\begin{split} \{a\}\\[10pt] (-\infty,\infty)\\[10pt] (a-R,a+R)\\[10pt] (a-R,a+R]\\[10pt] [a-R,a+R) \\[10pt] [a-R,a+R] \end{split}\]

Radius of Convergence

The Radius of Convergence is the largest number \(R\) such that for all \(x\)-values satisfying \(|x-a|<R\), the power series converges. Essentially, this gives us a measure of the size of the interval (it’s half of the width).

Example 2#

Find the radius of convergence and interval of convergence for power series:

\[ \sum_{n=0}^{\infty} \dfrac{(x-5)^n}{n^2} \]

Try for yourself and then click to see the steps.

We start by setting up the ratio test (normally we start with either the Ratio or the Root Test)

\[\begin{split} \left|\dfrac{a_{n+1}}{a_n}\right| &= \left|a_{n+1}\cdot \dfrac{1}{a_n}\right|\\[10pt] &= \left|\dfrac{(x-5)^{n+1}}{(n+1)^2} \cdot \dfrac{n^2}{(x-5)^n}\right|\\[10pt] &= \left|\dfrac{(x-5)^{n+1}}{(x-5)^n} \cdot \dfrac{n^2}{(n+1)^2}\right|\\[10pt] &= \left|(x-5) \cdot \left(\dfrac{n}{n+1}\right)^2\right|\\[10pt] &= \left|x-5\right| \cdot \left(\dfrac{n}{n+1}\right)^2\\[10pt] \end{split}\]

Note that since we don’t know what \(x\) is, the term \(x-5\) could potentially be negative. This means we can’t drop the absolute value for this \(x\)-term.

The next step is to calculate the limit for the ratio test:

\[ \lim_{n\to \infty} \left|\dfrac{a_{n+1}}{a_n}\right| = \lim_{n\to \infty} \left|x-5\right| \cdot \left(\dfrac{n}{n+1}\right)^2 \]

Since we are doing the limit as \(n\) goes to infinity, we are actually viewing \(x\) as a fixed constant with respect to this limit. This means we can actually move the absolute value term in front of the limit:

\[ = \left|x-5\right| \cdot \lim_{n\to \infty} \left(\dfrac{n}{n+1}\right)^2 \]

And then calculate the remaining limit, using all of our usual methods (continuous outer function and \(1/n\)-method):

\[\begin{split} & = \left|x-5\right| \cdot \left(\lim_{n\to \infty} \dfrac{n}{n+1}\right)^2\\[10pt] & = \left|x-5\right| \cdot \left(\lim_{n\to \infty} \dfrac{1}{1+\tfrac{1}{n}}\right)^2\\[10pt] & = \left|x-5\right| \cdot \left(\dfrac{1}{1+0}\right)^2\\[10pt] & = \left|x-5\right|\\[10pt] \end{split}\]

At this point, we are ready to determine our first-draft for the interval of convergence. This won’t be the final answer but it will be close.

By the Ratio Test:

  • if \(|x-5|<1\) the series converges.

  • if \(|x-5|>1\) the series diverges.

  • if \(|x-5|=1\) the test is inconclusive.

This means we can determine the interval except for the endpoints:

\[\begin{split} & \quad \qquad |x-5|<1 \\[10pt] &\iff -1<x-5<1 \\[10pt] & \iff 4<x<6 \\[10pt] \end{split}\]

So the interval of convergence is going to involve the interval \((4,6)\) but we still need to check the endpoints (the Ratio Test is inconclusive there).

At this point, we can determine the radius of convergence by looking at the width of the interval \((4,6)\)

The width of this interval is \(2\), so half of this is \(1\). (Think of circles, the radius is half the diameter.)

Therefore we say the radius of convergence is: \(R=1\).

To finalize our interval of convergence, we next need to check the endpoints. Because the ratio test is inconclusive at each one, we will need to use other tests besides the ratio and root tests.

Let’s start with endpoint: \(x=6\). Plugging this into the series (wherever we see an \(x\)) gives us:

\[ x=6 \quad \longrightarrow \quad \sum_{n=0}^{\infty} \dfrac{(6-5)^n}{n^2} = \sum_{n=0}^{\infty} \dfrac{1}{n^2} \]

which is a convergent \(p\)-series with \(p=2\) which satisfies \(p>1\).

Next we check the other endpoint: \(x=4\). Plugging this in gives us:

\[ x=4 \quad \longrightarrow \quad \sum_{n=0}^{\infty} \dfrac{(4-5)^n}{n^2} = \sum_{n=0}^{\infty} \dfrac{(-1)^n}{n^2} \]

which is convergent by the Absolute Convergence Test since \(\sum |a_n| = \sum\left| \tfrac{(-1)^n}{n^2}\right| = \sum \tfrac{1}{n^2}\) is a convergent \(p\)-series. (Note that we could also use the alternating series test here.)

Since the power series converges at both endpoints, we say that the interval of convergence is \([4,6]\).

Note that we use:

  • square brackets [ or ] to indicate that we want to include an endpoint (because it is convergent).

  • open parentheses ( or ) to indicate that we want to exclude an endpoint (because it is divergent).

Example 3#

Find the radius of convergence and interval of convergence for power series:

\[ \sum_{n=0}^{\infty} \dfrac{x^n}{(2n)!} \]

Try for yourself and then click to see the steps.

We start by setting up the ratio test (normally we start with either the Ratio or the Root Test)

\[\begin{split} \left|\dfrac{a_{n+1}}{a_n}\right| &= \left|a_{n+1}\cdot \dfrac{1}{a_n}\right|\\[10pt] &= \left|\dfrac{x^{n+1}}{(2(n+1))!} \cdot \dfrac{(2n)!}{x^n}\right|\\[10pt] &= \left|\dfrac{x^{n+1}}{x^n} \cdot \dfrac{(2n)!}{(2n+2)!}\right|\\[10pt] &= \left|x \cdot \dfrac{(2n)!}{(2n+2)(2n+1)(2n)!}\right|\\[10pt] &= \left|x\right| \cdot \dfrac{1}{(2n+2)(2n+1)}\\[10pt] \end{split}\]

Note that since we don’t know what \(x\) is, we can’t drop the absolute value for this \(x\)-term.

The next step is to calculate the limit for the ratio test:

\[ \lim_{n\to \infty} \left|\dfrac{a_{n+1}}{a_n}\right| = \lim_{n\to \infty} \left|x\right| \cdot \dfrac{1}{(2n+2)(2n+1)} = \left|x\right| \cdot \lim_{n\to \infty} \dfrac{1}{(2n+2)(2n+1)} = |x|\cdot 0 =0 \]

Since this is a \(\tfrac{c}{\infty}\)-form limit, which we know goes to zero.

For any \(x\)-value, we just saw that the limit is always 0. Therefore by the Ratio Test, the series converges for all \(x\)-values, and we say:

\[ \text{Interval of Convergence} \quad (-\infty,\infty) \]

Since we can think of the radius of convergence as a measure of the size of the interval (half of the width), we would say:

\[ \text{Radius of Convergence} \quad R=\infty \]

Example 4#

Find the radius of convergence and interval of convergence for power series:

\[ \sum_{n=0}^{\infty} \dfrac{(-1)^{n-1}3^n}{2n+1}(x-1)^n \]

Try for yourself and then click to see the steps.

We start by setting up the ratio test (normally we start with either the Ratio or the Root Test)

\[\begin{split} \left|\dfrac{a_{n+1}}{a_n}\right| &= \left|a_{n+1}\cdot \dfrac{1}{a_n}\right|\\[10pt] &= \left|\dfrac{(-1)^{n}3^{n+1}(x-1)^{n+1}}{2(n+1)+1} \cdot \dfrac{2n+1}{(-1)^{n-1}3^n(x-1)^n}\right|\\[10pt] &= \left|\dfrac{(-1)^{n}}{(-1)^{n-1}} \cdot \dfrac{2n+1}{2n+3}\cdot \dfrac{3^{n+1}}{3^n} \cdot \dfrac{(x-1)^{n+1}}{(x-1)^n}\right|\\[10pt] &= \left|\dfrac{2n+1}{2n+3}\cdot 3 \cdot (x-1)\right|\\[10pt] &= 3 \cdot \dfrac{2n+1}{2n+3}\cdot \left|x-1\right|\\[10pt] \end{split}\]

Note that since we don’t know what \(x\) is, the term \(x-1\) could potentially be negative. This means we can’t drop the absolute value for this \(x\)-term.

The next step is to calculate the limit for the ratio test:

\[ \lim_{n\to \infty} \left|\dfrac{a_{n+1}}{a_n}\right| = \lim_{n\to \infty} 3 \cdot \dfrac{2n+1}{2n+3}\cdot \left|x-1\right| \]

Since we are doing the limit as \(n\) goes to infinity, we are actually viewing \(x\) as a fixed constant with respect to this limit. This means we can actually move the absolute value term in front of the limit:

\[ = 3 \left|x-1\right| \cdot \lim_{n\to \infty} \dfrac{2n+1}{2n+3} \]

And then calculate the remaining limit, using all of our usual \(1/n\)-method:

\[\begin{split} & = 3 \left|x-1\right| \cdot \lim_{n\to \infty} \dfrac{2+\tfrac{1}{n}}{2+\tfrac{3}{n}}\\[10pt] & = 3 \left|x-1\right| \cdot \dfrac{2+0}{2+0}\\[10pt] & = 3 \left|x-1\right| \\[10pt] \end{split}\]

At this point, we are ready to determine our first-draft for the interval of convergence. This won’t be the final answer but it will be close.

By the Ratio Test:

  • if \(3|x-1|<1\) the series converges.

  • if \(3|x-1|>1\) the series diverges.

  • if \(3|x-1|=1\) the test is inconclusive.

This means we can determine the interval except for the endpoints:

\[\begin{split} & \quad \qquad 3|x-1|<1 \\[10pt] & \iff \quad |x-1|<\tfrac{1}{3} \\[10pt] & \iff -\tfrac{1}{3}<x-1<\tfrac{1}{3} \\[10pt] & \iff \tfrac{2}{3}<x<\tfrac{4}{3} \\[10pt] \end{split}\]

So the interval of convergence is going to involve the interval \(\left( \tfrac{2}{3},\tfrac{4}{3} \right)\) but we still need to check the endpoints (the Ratio Test is inconclusive there).

Now we can determine the radius of convergence by calculating the width of the interval \(\left( \tfrac{2}{3},\tfrac{4}{3} \right)\) to be \(\tfrac{2}{3}\). And then half of this is \(\tfrac{1}{3}\). (Think of circles, the radius is half the diameter.)

\[ \text{Radius of Convergence} \quad R=\dfrac{1}{3} \]

To finalize our interval of convergence, we next need to check the endpoints. Because the ratio test is inconclusive at each one, we will need to use other tests besides the ratio and root tests.

Let’s start with endpoint: \(x=\tfrac{4}{3}\). Plugging this into the series (wherever we see an \(x\)) gives us:

\[\begin{split} \sum_{n=0}^{\infty}\dfrac{(-1)^{n-1}3^n}{2n+1}\cdot \left(\tfrac{4}{3}-1\right)^n &= \sum_{n=0}^{\infty}\dfrac{(-1)^{n-1}3^n}{2n+1}\cdot \left(\tfrac{1}{3}\right)^n\\[10pt] & = \sum_{n=0}^{\infty}\dfrac{(-1)^{n-1}}{2n+1}\\[10pt] \end{split}\]

This is convergent by the Alternating Series Test (show this).

Next we check the other endpoint: \(x=\tfrac{2}{3}\). Plugging this in gives us:

\[\begin{split} \sum_{n=0}^{\infty}\dfrac{(-1)^{n-1}3^n}{2n+1}\cdot \left(\tfrac{2}{3}-1\right)^n & = \sum_{n=0}^{\infty}\dfrac{(-1)^{n-1}3^n}{2n+1}\cdot \left(-\tfrac{1}{3}\right)^n \\[10pt] & = \sum_{n=0}^{\infty}\dfrac{(-1)^{n-1}3^n}{2n+1}\cdot (-1)^n\left(\tfrac{1}{3}\right)^n \\[10pt] & = \sum_{n=0}^{\infty}\dfrac{-1}{2n+1} \\[10pt] & = -\sum_{n=0}^{\infty}\dfrac{1}{2n+1} \\[10pt] \end{split}\]

which can be shown to be divergent by the Limit Comparison Test (show this).

Since the power series diverges at both endpoints, we say:

\[ \text{Interval of Convergence} \quad \left( \tfrac{2}{3},\tfrac{4}{3} \right) \]

Note that we use:

  • square brackets [ or ] to indicate that we want to include an endpoint (because it is convergent).

  • open parentheses ( or ) to indicate that we want to exclude an endpoint (because it is divergent).

Example 5#

Find the radius of convergence and interval of convergence for power series:

\[ \sum_{n=0}^{\infty} n! (x-3)^n \]

Try for yourself and then click to see the steps.

We start by setting up the ratio test (normally we start with either the Ratio or the Root Test)

\[\begin{split} \left|\dfrac{a_{n+1}}{a_n}\right| &= \left|a_{n+1}\cdot \dfrac{1}{a_n}\right|\\[10pt] &= \left|\dfrac{(n+1)! (x-3)^{n+1}}{n! (x-3)^n} \right|\\[10pt] &= \left|\dfrac{(n+1)n! (x-3)^{n+1}}{n! (x-3)^n} \right|\\[10pt] &= \left| (n+1)(x-3) \right|\\[10pt] &= (n+1)|x-3|\\[10pt] \end{split}\]

Note that since we don’t know what \(x\) is, we can’t drop the absolute value for this \(x\)-term.

The next step is to calculate the limit for the ratio test:

\[ \lim_{n\to \infty} \left|\dfrac{a_{n+1}}{a_n}\right| = \lim_{n\to \infty} (n+1)|x-3| = \left|x-3 \right| \cdot \lim_{n\to \infty} (n+1) = \infty \]

There is a technical point here:

  • The Ratio Test is applicable for all \(x\)-values except the center value of our power series. For this problem, the center value is \(a=3\).

This means that when we calculated the above limit, it was actually for \(x\neq 3\) which is equivalent to \(|x-3|\neq 0\).

Therefore, by the Ratio Test, the power series diverges for all \(x\neq 3\).

For \(x\neq 3\), we found the power series diverges.

What about when \(x=3\)? Since this is the center of our power series, we are guaranteed to have convergence here.

Therefore, this power series only converges at \(x=3\), so we have a collapsed interval and write:

\[ \text{Interval of Convergence} \quad \{3\}=[3,3] \]

To see why the power series converges at the center value, let’s plug \(x=3\) into the power series:

\[\begin{split} \sum_{n=0}^{\infty} n! (3-3)^n &= \sum_{n=0}^{\infty} n! \cdot (0)^n \\[10pt] &= 0!\cdot 0^0+1!\cdot 0^1+ 2!\cdot 0^2+3!\cdot 0^3+\cdots \\[10pt] & =1+0+ 0+0+\cdots \\[10pt] & = 1 \end{split}\]

Essentially, we end up adding more and more copies of 0 which we can always do, their sum is always 0.

One comment about the notation: \(0!=1\) and by convention \(0^0=1\).

Since we can think of the radius of convergence as a measure of the size of the interval (half of the width), we would say:

\[ \text{Radius of Convergence} \quad R=0 \]

Since our interval has 0 width, the radius is 0 as well.

Example 6#

Find the radius of convergence and interval of convergence for power series:

\[ \sum_{n=0}^{\infty} \dfrac{n(x+1)^n}{2^{n+1}} \]

Try for yourself and then click to see the steps.

We start by setting up the ratio test (normally we start with either the Ratio or the Root Test)

\[\begin{split} \left|\dfrac{a_{n+1}}{a_n}\right| &= \left|a_{n+1}\cdot \dfrac{1}{a_n}\right|\\[10pt] &= \left|\dfrac{(n+1)(x+1)^{n+1}}{2^{n+2}} \cdot \dfrac{2^{n+1}}{n(x+1)^n}\right|\\[10pt] &= \left|\dfrac{n+1}{n} \cdot \dfrac{(x+1)^{n+1}}{(x+1)^{n}}\cdot \dfrac{2^{n+1}}{2^{n+2}} \right|\\[10pt] &= \left|\dfrac{n+1}{n} \cdot (x+1)\cdot \dfrac{1}{2} \right|\\[10pt] &= \tfrac{1}{2} \cdot \dfrac{n+1}{n}\cdot \left|x+1\right|\\[10pt] \end{split}\]

Note that since we don’t know what \(x\) is, the term \(x+1\) could potentially be negative. This means we can’t drop the absolute value for this \(x\)-term.

The next step is to calculate the limit for the ratio test:

\[ \lim_{n\to \infty} \left|\dfrac{a_{n+1}}{a_n}\right| = \lim_{n\to \infty} \tfrac{1}{2} \cdot \dfrac{n+1}{n}\cdot \left|x+1\right| \]

Since we are doing the limit as \(n\) goes to infinity, we are actually viewing \(x\) as a fixed constant with respect to this limit. This means we can actually move the absolute value term in front of the limit, and calculate:

\[\begin{split} &= \tfrac{1}{2} \left|x+1\right| \cdot \lim_{n\to \infty} \dfrac{n+1}{n}\\[10pt] &= \tfrac{1}{2} \left|x+1\right| \cdot \lim_{n\to \infty} \left( 1 +\dfrac{1}{n}\right)\\[10pt] &= \tfrac{1}{2} \left|x+1\right| \cdot \left( 1 +0\right)\\[10pt] &= \tfrac{1}{2} \left|x+1\right|\\[10pt] \end{split}\]

At this point, we are ready to determine our first-draft for the interval of convergence. This won’t necessarily be the final answer but it will be close.

By the Ratio Test:

  • if \(\tfrac{1}{2} \left|x+1\right|<1\) the series converges.

  • if \(\tfrac{1}{2} \left|x+1\right|>1\) the series diverges.

  • if \(\tfrac{1}{2} \left|x+1\right|=1\) the test is inconclusive.

This means we can determine the interval except for the endpoints:

\[\begin{split} & \quad \qquad \tfrac{1}{2} \left|x+1\right|<1 \\[10pt] & \iff \quad |x+1|<2 \\[10pt] & \iff -3<x+1<1 \\[10pt] & \iff -3<x<1 \\[10pt] \end{split}\]

So the interval of convergence is going to involve the interval \(\left( -3, 1 \right)\) but we still need to check the endpoints (the Ratio Test is inconclusive there).

Now we can determine the radius of convergence by calculating the width of the interval \(\left( -3, 1 \right)\) to be \(4\). And then half of this is \(2\). (Think of circles, the radius is half the diameter.)

\[ \text{Radius of Convergence} \quad R=2 \]

To finalize our interval of convergence, we next need to check the endpoints. Because the ratio test is inconclusive at each one, we will need to use other tests besides the ratio and root tests.

Let’s start with endpoint: \(x=1\). Plugging this into the series (wherever we see an \(x\)) gives us:

\[\begin{split} \sum_{n=0}^{\infty} \dfrac{n(1+1)^n}{2^{n+1}} &= \sum_{n=0}^{\infty} \dfrac{n 2^n}{2^{n+1}}\\[10pt] & = \sum_{n=0}^{\infty}\dfrac{n}{2}\\[10pt] \end{split}\]

This is divergent by the Test for Divergence (show this).

Next we check the other endpoint: \(x=-3\). Plugging this in gives us:

\[\begin{split} \sum_{n=0}^{\infty} \dfrac{n(-3+1)^n}{2^{n+1}} &= \sum_{n=0}^{\infty} \dfrac{n(-2)^n}{2^{n+1}}\\[10pt] &= \sum_{n=0}^{\infty} \dfrac{n(-1)^n 2^n}{2^{n+1}}\\[10pt] & = \sum_{n=0}^{\infty}(-1)^n\dfrac{n}{2}\\[10pt] \end{split}\]

This is divergent by the Test for Divergence (show this).

Since the power series diverges at both endpoints, we say:

\[ \text{Interval of Convergence} \quad \left( -3,1 \right) \]

Note that we use:

  • square brackets [ or ] to indicate that we want to include an endpoint (because it is convergent).

  • open parentheses ( or ) to indicate that we want to exclude an endpoint (because it is divergent).

Summary#

Example

Series

Interval of Convergence

Radius of Convergence

Ex. 1

\(\displaystyle \sum_{n=0}^{\infty} x^n \)

\((-1,1)\)

\(R=1\)

Ex. 2

\(\displaystyle \sum_{n=0}^{\infty} \dfrac{(x-5)^n}{n^2} \)

\([4,6]\)

\(R=1\)

Ex. 3

\(\displaystyle \sum_{n=0}^{\infty} \dfrac{x^n}{(2n)!} \)

\((-\infty,\infty)\)

\(R=\infty\)

Ex. 4

\(\displaystyle \sum_{n=0}^{\infty} \dfrac{(-1)^{n-1}3^n}{2n+1}(x-1)^n \)

\(\left(\tfrac{2}{3},\tfrac{4}{3}\right]\)

\(R=\tfrac{1}{3}\)

Ex. 5

\(\displaystyle \sum_{n=0}^{\infty} n! (x-3)^n \)

\(\{3\}\)

\(R=0\)

Ex. 6

\(\displaystyle \sum_{n=0}^{\infty} \dfrac{n(x+1)^n}{2^{n+1}} \)

\((-3,1)\)

\(R=2\)